Question

1. A commercial prawn fisherman recorded the number of prawns he took from each trap. The average
number of prawns per trap was 230 with a standard deviation of 22. What number of prawns per trap
would you expect in the interval symmetrical about the mean where 80% of the numbers would be
found?

Answer 1

The numbers would be between 10% and 90% mark, do you know how to find z-scores?

Answer 2

The equation:
\[z = (x - \mu)/\sigma \]

Answer 3

idk how to find the z scores
i know they are -1.28 and 1.28

Answer 4

but idk how to get them

Answer 5

-1.28 is incorrect, you get them from z- tables http://www.z-table.com/

Answer 6

how do i know which score it is from the table

Answer 7

10% is less than mean, so to the left, hence is negative
90% is positive for same reason

Answer 8

You have \[\mu = 230, \sigma = 22, z - two different scenarios\]
You need to find out x, there will be two of them, one for minimum and one for maximum mark

Answer 9

Can you find x based on what you have?

Answer 10

how did u get -2.33 and 1.28

Answer 11

im not sure how to read thetable

Answer 12

See the attached, here I highlighted the one for 10%. By the way you are correct, they are -1.28 and 1.28

Answer 13

The closest number to 10% or 0,1 is 0,1003

Answer 14

See if you get it

Answer 15

oh okay i undestand now that part now thank you

Answer 16

so then z=x-m/o

Answer 17

to find x?

Answer 18

Yes, get two values of x

Answer 19

1.28 = (x - 230)/22
-1.28 = (x - 230)/22
Solve each for x

Answer 20

What did you get?

Answer 21

Any help?

Answer 22

thanks i got 201 to 258

Answer 23

Yes, I believe it is correct

Answer 24

thanks for your help

Answer 25

No problem